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Why do photons have momentum yet no mass?

  1. Mar 7, 2013 #1
    I imagine this gets asked a lot but i'm looking for an in depth explanation since none of the others I've found are detailed enough.

    I understand that p = mv is not applicable near the speed of light and I think (but am unsure) that the correct equation to use is e = pc.

    Assuming that e = pc is the correct equation to use then what does 'p' represent?

    Assuming e = pc is the wrong equation then what is the right one?

    And can someone (if you can be bothered to waste time on a dumbass like me) explain why p = mv is not applicable?

    Thanks.
     
  2. jcsd
  3. Mar 7, 2013 #2
    No one knows 'why' a photon exists nor why it has the characteristics it does....
    the fundamental particles of this universe, as well as mass, energy, time, distance,etc....were all 'unified' [combined] in one entity very early in our universe....that was a very high energy unstable environment which underwent what is called 'spontaneous symmetry breaking'....meaning the original entity broke down to a lower energy state and became all those different entities I listed...... which is what we observe today.

    with p = mv, what would you use the the 'm' of a photon??

    For a correct description, see here:

    http://en.wikipedia.org/wiki/Photon

    Unlike a matter particle, a photon always moves in a vacuum at 'c' and if a photon has higher energy it has higher frequency....so a gamma ray, for example, is very energetic [high frequency] visible light less so.
     
  4. Mar 7, 2013 #3

    ZapperZ

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    Please note that even in the classical treatment of light as EM wave and not photons, there is still a momentum associated with light! This is a standard treatment of classical E&M. So one does not need to have a picture of massless photons to already realize that light can have a momentum.

    Zz.
     
  5. Mar 7, 2013 #4

    jtbell

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    The relativistic relationship between a particle's energy, momentum and mass is

    $$E^2 = (pc)^2 + (mc^2)^2$$

    where m is what many people call "rest mass" but physicists usually call just "mass." Setting m = 0 (as for a photon) gives you E = pc.
     
  6. Mar 7, 2013 #5
    It IS applicable NEAR but not AT the speed of light....

    edit: sloppy explanation by me...see last line of the next post..by Bill_K..
     
    Last edited: Mar 7, 2013
  7. Mar 7, 2013 #6

    Bill_K

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    Only if you mean m to be the antiquated concept of relativistic mass. In terms of the usual rest mass, p = γmv.
     
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